Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{-5r + 45}{-2r^2 + 22r - 36} \times \dfrac{-r^2 + 3r - 2}{4r + 8} $
First factor out any common factors. $x = \dfrac{-5(r - 9)}{-2(r^2 - 11r + 18)} \times \dfrac{-(r^2 - 3r + 2)}{4(r + 2)} $ Then factor the quadratic expressions. $x = \dfrac {-5(r - 9)} {-2(r - 2)(r - 9)} \times \dfrac {-(r - 2)(r - 1)} {4(r + 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {-5(r - 9) \times -(r - 2)(r - 1) } { -2(r - 2)(r - 9) \times 4(r + 2)} $ $x = \dfrac {5(r - 2)(r - 1)(r - 9)} {-8(r - 2)(r - 9)(r + 2)} $ Notice that $(r - 2)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {5\cancel{(r - 2)}(r - 1)(r - 9)} {-8\cancel{(r - 2)}(r - 9)(r + 2)} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $x = \dfrac {5\cancel{(r - 2)}(r - 1)\cancel{(r - 9)}} {-8\cancel{(r - 2)}\cancel{(r - 9)}(r + 2)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $x = \dfrac {5(r - 1)} {-8(r + 2)} $ $ x = \dfrac{-5(r - 1)}{8(r + 2)}; r \neq 2; r \neq 9 $